Question

I NEED HELP

Answer 1

Given:

The graph of a sine function.

To find:

The sine function.

Solution:

The general form of a sine function is

$y=Asin(Bx+C)+D$            ...(i)

Where, |A| is amplitude, $\dfrac{2\pi}{B}$ is period, $-\dfrac{C}{B}$ is phase shift and D is mid-line.

From the given graph it is clear that the minimum value of the function is -4 and the maximum value is 0.

$|A|=\dfrac{Maximum-Minimum}{2}$

$|A|=\dfrac{0-(-4)}{2}$

$|A|=\dfrac{4}{2}$

$|A|=2$

The graph is reflected across the mid-line, so A=-2.

And,

$D=\dfrac{Maximum+Minimum}{2}$

$D=\dfrac{0+(-4)}{2}$

$D=\dfrac{-4}{2}$

$D=-2$

Period of the function is 4π because it completer its one cycle in the interval of 4π.

$4\pi=\dfrac{2\pi}{B}$

$B=\dfrac{2\pi}{4\pi}$

$B=\dfrac{1}{2}$

Phase shift is $\dfrac{\pi}{4}$.

$-\dfrac{C}{B}=\dfrac{\pi}{4}$

$-\dfrac{C}{\frac{1}{2}}=\dfrac{\pi}{4}$

$C=-\dfrac{1}{2}\times \dfrac{\pi}{4}$

$C=-\dfrac{\pi}{8}$

Putting $A=-2,B=\dfrac{1}{2},C=-\dfrac{\pi}{8},D=-2$ in (i), we get

$y=-2\sin\left(\dfrac{1}{2}x-\dfrac{\pi}{8}\right)-2$

Therefore, the required equation is $y=-2\sin\left(\dfrac{1}{2}x-\dfrac{\pi}{8}\right)-2$.